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Ilia_Sergeevich [38]
3 years ago
5

A gas is allowed to expand from a volume of 400 ml to 2000 ml at a constant temperature. If the the initial pressure is 3 atm, c

alculate the final pressure exerted by the gas
Chemistry
1 answer:
Lostsunrise [7]3 years ago
8 0

Answer:

P_{2}) = 0.6 \; atm

Explanation:

<u>Given the following data;</u>

Initial volume = 400 mL

Final volume = 2000 mL

Initial pressure = 3 atm

To find the final pressure P2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

PV = K

P_{1}V_{1} = P_{2}V_{2}

Substituting into the equation, we have;

3 * 400 = P_{2}* 2000

1200 = 2000P_{2}

P_{2}) = \frac {1200}{2000}

P_{2}) = 0.6 \; atm

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0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was
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Answer: The value of k_{a} for the weak acid is 3.07 \times 10^{-4}.

Explanation:

First, we will calculate the molarity of HA as follows.

     [HA] = \frac{\text{no. of moles}}{\text{volume}}

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             = 0.015

At equilibrium,

              HA + H_{2}O \rightleftharpoons H_{3}O^{+} + A^{-}

Initial:   0.015                  0          0

Change:  -x                     +x         +x

Equilibm: 0.015 - x          x           x

It is given that, at equilibrium

          [HA] = 0.015 - x = 0.013

             - x = 0.013 - 0.015

                  = 0.002

Now, expression for k_{a} of this reaction is as follows.

          k_{a} = \frac{[A^{-}][H_{3}O^{+}]}{[HA]}

                      = \frac{x^{2}}{[HA]}

                      = \frac{(0.002)^{2}}{0.013}

                      = 3.07 \times 10^{-4}

Thus, we can conclude that the value of k_{a} for the weak acid is 3.07 \times 10^{-4}.

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